1. Let ϕ be a homomorphism from G to H and let K be the kernel of ϕ. Claim the set of subsets of the form ak forms a group.

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Millicent Ward
5 years ago
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1 Math 541 Day Let ϕ be a homomorphism from G to H and let K be the kernel of ϕ. Claim the set of subsets of the form ak forms a group. What is the operation? Suppose we want to multiply ak and bk. How would we do that? What would we get for an answer? Answer: We could do it by multiplying everything in ak by everything in bk and see what we come up with. So, let s consider the set akbk = {akbk : k, k is in K}. We want this to be equal to some set of the form gk. The logical candidate is abk. Let s see if we can prove that abk = akbk. Suppose g is in abk, then g = (ab)k for some k in K. Then g = (aeg)(bk) where eg is the identity element of G. Note that eg is in K, so g is in akbk. This proves that abk akbk We ll have to be a bit more clever to prove the other containment! Suppose that g is in akbk, then g = akbk for some k, k in K. If we could commute k and b, we would be golden (do you see why?). Well we can t do that, but we can sort of commute them using the fact that bk = Kb. First, why is that true? Quick way to see it is true: We know that bk is exactly the set of elements that maps to ϕ(b) [See first item on Day 20]. We also know that Kb is exactly the set of elements that maps to ϕ(b) [ You proved it in exercise 52 of Section 13, and it is essentially the same proof that I did on Day 20]. So these must be the same set. We can also prove it directly: Suppose that g is in bk, then g = bk for some k in K. This means ϕ(g) = ϕ(bk) = ϕ(b) ϕ(k) = ϕ(b) eg = ϕ(b). But then we know that ϕ(g)ϕ(b) - 1 = eg. Which means that ϕ(gb - 1 ) = eg, so that gb - 1 is in K. Thus gb - 1 = k for some k in K. So that g = k b is in Kb. The proof of the other containment is very similar.

2 This means that anything of the form bk where k is in K can be written as k b for some (possibly different) k in K. So the expression akbk can be rewritten as abk k for some k in K, proving that g = akbk is an element of abk. So that is awesome, we can now try to define the operation by akbk = abk. How do we know this is an operation? Is it everywhere defined? Is it well- defined? It is clearly everywhere defined because given two cosets, ak, bk, our formula gives us a coset for an answer. Well- defined is more of a problem because a coset can have multiple names. For example: if k is in the kernel, then kk is certainly the same as ek because K is a subgroup (so is closed under multiplication). More generally, if ϕ(g) = ϕ(a) then gk = ak because both of these are exactly the set of elements that map to whatever ϕ(g) = ϕ(a) is. NOTE: You should also be able to prove directly that ϕ(g) = ϕ(a) then gk = ak and should try to do so! So to prove well- defind, we need to suppose that ak = ck and bk = dk and prove that when we multiply these cosets, we get the same answer no matter which version of each coset we use. So we need to prove that akbk = abk is the same as ckdk = cdk. So how do we show abk = cdk? This is going to require a set equality proof unless we can think of a slicker way to prove two cosets are equal. Here is where we want the Handy Little Coset Lemma. It goes like this:

3 Handy Little Coset Lemma (HLCL): For any subgroup H of a group G, and any pair of elements a, b in G, ah = bh if and only if a - 1 b is in H. Proof: Suppose ah = bh. First observe that since b is clearly in bh, we know that b is in ah. This means that b = ah for some h in H. This means that a - 1 b = h is in H. Now suppose a - 1 b is in H and let x in ah. Then x = ah for some h in H. But since a - 1 b is in H, so is its inverse b - 1 a. This means that a = bh for some h in H. So know we have x = bh h which is in bh. If we let x in bh, we have x = bh for some h in H. But a - 1 b is in H, so b = ah for some h in H. So x = ah h which is in ah. So we have shown that ah = bh. Thus HLCL has been proven and is now available for use! Now we can show that the operation given by akbk = abk is well defined: Suppose that ak = ck and bk = dk. We need to prove that akbk = abk is the same as ckdk = cdk. By HLCL, abk = cdk iff (ab) - 1 cd = b - 1 a - 1 cd is in K. However, (also by HLCL) we already know that a - 1 c and b - 1 d are in K. So ϕ(b - 1 a - 1 cd) = ϕ(b - 1 ) ϕ(a - 1 c) ϕ(d) = ϕ(b - 1 )ehϕ(d) = ϕ(b - 1 )ϕ(d) = ϕ(b - 1 d) = eh Thus, (ab) - 1 cd is in K and so abk = cdk. This means our operation is well defined! Notation: G/K means the set of left cosets of the kernel. Formally, G/K ={gk : g is in G}. The set G/K with the operation given by akbk = abk forms a group. The group properties follow immediately since the operation is defined in terms of the operation on G. egk = K is the identity coset. For any coset ak, a - 1 K is its inverse. Associativity: (akbk)ck = (ab)kck = [(ab)c]k = [a(bc)]k = ak(bc)k = ak(bkck).

4 Now that we know that G/K is a group w.r.t. coset multiplication, we are only one step away from being able to prove the fundamental homomorphism theorem. Claim: The natural (or canonical) map γ: G G/K given by γ (g) = gk is a homomorphism. Proof: Easy practice with the definition of homomorphism and the definition of coset multiplication. [Left to you] The Fundamental Homomorphism Theorem: Let ϕ be a homomorphism from G to H and let K be the kernel of ϕ. Then the map µ: G/K ϕ(g) given by µ(gk) = ϕ(g) is an isomorphism. If we letγ: G G/K be the natural homomorphism, then ϕ = µoγ. Proof: First will show µ is an isomorphism. Onto: Let ϕ(g) be in ϕ(g), then µ(gk) = ϕ(g). 1-1: Suppose µ(ak) = µ(bk). Then ϕ(a) = ϕ(b). But then ϕ(a - 1 b) = eh. This means a - 1 b is in K. So by the HLCL we know ak = bk. Homomorphism: µ(akbk) = µ(abk) = ϕ(ab) = ϕ(a) ϕ(b) = µ(ak) µ(bk). [Note that the above is just formally capturing 2- step process of turning a homomorphism into an isomorphism.] The last bit is even easier: For all g in G, µoγ(g) = µ(γ(g)) =µ(gk) = ϕ(g).

5 Example: Consider ϕ: Z D8 given by ϕ(n) = R n. It is a homomorphism because ϕ(n+m) = R n+m = R n R m = ϕ(n) ϕ(m). The kernel is 4Z since for any k in Z, ϕ(4k)= R 4k =(R 4 ) k = I. The image is {I, R, R 2, R 3 }. The Fundamental Homomorphism Theorem gives us an isomorphism from Z/4Z to {I, R, R 2, R 3 } given by 4Z 4Z+1 R 4Z+2 R 2 4Z+3 R 3 I Note that we are using additive notation so ak becomes a+k. Further, in this case we usually use right cosets instead of left. Since Z is commutative this is no problem at all.

6 Notice that this tells us that Z/4Z is a cyclic group of order 4 and so it is isomorphic to Z 4. By a similar approach any Z/nZ is isomorphic to Z n. In fact this is often how Z n is defined. In this case, the notation is often simplified so that, for example, 4Z+3 is expressed as [3] or [3]4 or sometimes as 3. Preview: We have seen that we can make a group out of cosets of the kernel of a homomorphism. But we can make cosets out of any subgroup. Next time we will figure out when these more general cosets can form a group.

Math 4/541 Day 25. Some observations:

Math 4/541 Day 25. Some observations: Math 4/541 Day 25 1. Previously we showed that given a homomorphism, ϕ, the set of (left) cosets, G/K of the kernel formed a group under the operation akbk = abk. Some observations: We could have just